Gas Laws is the most algebra-heavy topic in Theme B of IB Physics. Every Paper 1 features at least one multiple-choice question on $PV = nRT$ or its particle-form cousin $PV = N k_B T$, and Paper 2 routinely asks for an extended energy-balance calculation that mixes kinetic theory with the ideal gas law. The kinetic-model assumptions also sit at the heart of any Paper 3 essay on thermodynamics or astrophysics.
This cheatsheet condenses every formula, definition, trick, and trap from Topic B.3 into one revisable page — kinetic model, the three empirical laws, the unified ideal gas law, $\bar{E}_k = \tfrac{3}{2} k_B T$, and the real-vs-ideal comparison. The worked example walks through a Boyle/Charles fusion problem in IB mark-scheme rhythm, and the FAQ section drills the conceptual misconceptions our students hit most often in mock exams.
§1 — Pressure & Kinetic Model Topic B.3
Key formulae
Pressure:$P = \dfrac{F}{A}$
Kinetic-theory $P$:$P = \tfrac{1}{3}\rho \langle v^2 \rangle$
rms speed:$v_\text{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{\dfrac{3 k_B T}{m}}$
Kinetic-model assumptions
- A large number of identical particles in constant random motion.
- Particle volume is negligible compared to the container.
- Collisions are perfectly elastic (KE conserved).
- No intermolecular forces between particles.
- Collision times are negligible compared to time between collisions.
- All directions of motion are equally probable.
TrickThe factor $\tfrac{1}{3}$ in $P = \tfrac{1}{3}\rho \langle v^2 \rangle$ comes from 3D random motion — only one third of molecules contribute to each direction at a time.
TrapPressure counts only forces perpendicular to the wall area. Tangential (parallel) forces don't generate pressure.
§2 — Amount of Substance Topic B.3
Key formulae
Moles from $N$:$n = \dfrac{N}{N_A}$
Moles from mass:$n = \dfrac{m}{M_r}$
Number from mass:$N = \dfrac{m N_A}{M_r}$
Boltzmann from R:$k_B = \dfrac{R}{N_A} = 1.38 \times 10^{-23}\,\text{J K}^{-1}$
Constants you must memorise
- $N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}$ — Avogadro constant
- $R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ — molar gas constant
- $k_B = 1.38 \times 10^{-23}\,\text{J K}^{-1}$ — Boltzmann constant
NoteTo find which gas sample (from equal masses) has the most molecules, pick the one with the smallest molar mass $M_r$ — smallest $M_r$ ⇒ most moles ⇒ most molecules.
§3 — Empirical Gas Laws Topic B.3
The three classical laws
| Law | Constant | Relation | Form |
|---|---|---|---|
| Boyle | $T$ | $P \propto 1/V$ | $P_1 V_1 = P_2 V_2$ |
| Charles | $P$ | $V \propto T$ | $V_1/T_1 = V_2/T_2$ |
| Gay-Lussac | $V$ | $P \propto T$ | $P_1/T_1 = P_2/T_2$ |
TrapTemperature must be in kelvin: $T(\text{K}) = \theta({}^\circ\text{C}) + 273$. Never substitute Celsius directly into a gas-law ratio.
TrickCombined gas law for any change of state with fixed mass: $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$. Cancel whichever variable stays constant.
Molecular explanations
- ↓V at constant T ⇒ ↑P: molecules hit walls more often (shorter distance between collisions).
- ↑T at constant V ⇒ ↑P: greater speed ⇒ more collisions per second and greater momentum transfer.
- ↑T at constant P ⇒ ↑V: greater speed ⇒ volume must expand to restore the original collision rate per area.
§4 — Ideal Gas Law Topic B.3
Three equivalent forms
Molar form:$PV = nRT$
Particle form:$PV = N k_B T$
Combined states:$\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$
Equivalence:$nR = N k_B \;\Leftrightarrow\; k_B = R/N_A$
TrickUse $PV = nRT$ when given mass or moles. Use $PV = N k_B T$ when given the number of particles. They give identical results — pick the form that matches the data.
Trap$PV/T = nR$ is constant only for fixed mass. If gas is added or escapes, $n$ changes and the ratio is no longer constant.
P–V diagram features
- Isothermal: hyperbola, $PV = $ constant; higher curve = higher $T$.
- Isobaric: horizontal line (constant $P$).
- Isochoric: vertical line (constant $V$).
- $P$ vs $1/V$ at constant $T$: straight line through origin.
§5 — Internal Energy & Average KE Topic B.3
Key formulae
Avg KE / molecule:$\langle E_K \rangle = \tfrac{3}{2} k_B T$
Internal energy (monatomic):$U = \tfrac{3}{2} N k_B T = \tfrac{3}{2} n R T$
rms speed:$v_\text{rms} = \sqrt{\dfrac{3 k_B T}{m}}$
Trick$\langle E_K \rangle = \tfrac{3}{2} k_B T$ depends on temperature only — independent of gas type. Equal $T$ ⇒ equal average KE per molecule, regardless of which gas.
NoteFor a monatomic ideal gas, $U$ is purely kinetic — no PE, no rotational or vibrational energy. So $U \propto T$ exactly, with no offset.
TrapHeavier molecules at the same $T$ move slower, not faster. From $\tfrac{1}{2} m v_\text{rms}^2 = \tfrac{3}{2} k_B T$, larger $m$ gives smaller $v_\text{rms}$.
§6 — Ideal vs Real Gases Topic B.3
Comparison table
| Ideal | Real | |
|---|---|---|
| Molecular volume | negligible | finite |
| Intermolecular forces | none | attractive / repulsive |
| $PV = nRT$ | exact | approximate |
| Compressibility $Z = PV/nRT$ | $= 1$ | $\neq 1$ |
| Internal energy | KE only | KE + PE |
| Condensation | never | possible |
Ideal approximation valid when…
- Low pressure — molecules far apart; finite volume and forces become negligible.
- High / moderate temperature — KE $\gg$ intermolecular PE; far from condensation.
- Far from the gas's boiling point.
TrapAt high $P$: $Z > 1$ (repulsive forces and finite molecular volume dominate). At low $T$ with moderate $P$: $Z < 1$ (attractive forces pull molecules together, lowering pressure below the ideal value).
TrickOn a real-gas P–V diagram, a flat (constant-$P$) section of an isotherm represents a phase change (condensation). An ideal-gas isotherm is a smooth hyperbola — no flat region.
§7 — Exam Attack Plan All sections
When you see this in the question — reach for that:
| Question trigger | Reach for |
|---|---|
| Pressure changes at constant temperature | Boyle: $P_1 V_1 = P_2 V_2$ |
| Volume vs temperature at constant pressure | Charles: $V_1/T_1 = V_2/T_2$ |
| Pressure vs temperature at constant volume | Gay-Lussac: $P_1/T_1 = P_2/T_2$ |
| Two of $P$, $V$, $T$ change | Combined: $P_1 V_1/T_1 = P_2 V_2/T_2$ |
| Mass / moles given | $PV = nRT$, $R = 8.31$ |
| Number of particles given | $PV = N k_B T$, $k_B = 1.38\times 10^{-23}$ |
| "Average KE per molecule" | $\langle E_K \rangle = \tfrac{3}{2} k_B T$ |
| "rms speed" | $v_\text{rms} = \sqrt{3 k_B T / m}$ |
| "Why is the gas not ideal?" | Quote: finite molecular volume + non-zero intermolecular forces |
| P–V diagram with flat region | Condensation — substance crossing phase boundary |
Worked Example — IB-Style Combined Gas Law
Question (HL Paper 2 style — 6 marks)
A sealed cylinder contains $0.020\,\text{mol}$ of an ideal monatomic gas at $T_1 = 27\,^\circ\text{C}$, pressure $P_1 = 1.0 \times 10^{5}\,\text{Pa}$ and volume $V_1$. The gas is heated to $T_2 = 327\,^\circ\text{C}$ while a piston allows the volume to expand isobarically. (a) Find $V_1$. (b) Find the new volume $V_2$. (c) Find the change in internal energy of the gas.
Solution
- Convert to kelvin: $T_1 = 300\,\text{K}$, $T_2 = 600\,\text{K}$. (R1)
- Apply $PV = nRT$ at state 1: $V_1 = nRT_1/P_1 = 0.020 \times 8.31 \times 300 / (1.0 \times 10^5)$. (M1)
- Evaluate: $V_1 \approx 4.99 \times 10^{-4}\,\text{m}^3 \approx 5.0 \times 10^{-4}\,\text{m}^3$. (A1)
- Isobaric process ⇒ $V/T$ constant: $V_2 = V_1 \times T_2/T_1 = 5.0 \times 10^{-4} \times 600/300 = 1.0 \times 10^{-3}\,\text{m}^3$. (A1)
- Monatomic ideal gas: $\Delta U = \tfrac{3}{2} nR \Delta T = \tfrac{3}{2}(0.020)(8.31)(300)$. (M1)
- Evaluate: $\Delta U \approx 74.8\,\text{J} \approx 75\,\text{J}$. (A1)
Examiner's note: The biggest losers in this question type forget to convert to kelvin, then back-substitute Celsius into the ratio $V_2/T_2 = V_1/T_1$ — answers come out negative, with no physical meaning. Always convert temperatures first, then chase the algebra.
Common Student Questions
Why must temperature be in kelvin for gas laws?
All the gas laws (Boyle, Charles, Gay-Lussac, ideal gas) are derived from extrapolating $P$ or $V$ vs $T$ to absolute zero, which sits at 0 K. Substituting Celsius gives nonsensical results — for example, a temperature of $0\,^\circ\text{C}$ would imply zero pressure or volume. Always convert with $T(\text{K}) = \theta({}^\circ\text{C}) + 273$ before substituting into any ratio or product.
When should I use $PV = nRT$ vs $PV = N k_B T$?
Use $PV = nRT$ when the problem gives mass or moles ($n$), with $R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$. Use $PV = N k_B T$ when the problem gives the number of molecules ($N$), with $k_B = 1.38 \times 10^{-23}\,\text{J K}^{-1}$. They are mathematically identical because $nR = N k_B$. Pick the form that matches the data — converting wastes time and risks rounding errors.
Do heavier gas molecules at the same temperature move faster?
No — heavier molecules move slower at the same temperature. From $\tfrac{1}{2} m v_\text{rms}^2 = \tfrac{3}{2} k_B T$, a larger $m$ gives a smaller $v_\text{rms}$. At equal $T$, all gases share the same average kinetic energy per molecule, but the heavier ones are slower. This is why oxygen molecules in air move slower than helium atoms at the same temperature.
When does the ideal gas approximation break down?
At high pressure, finite molecular volume and intermolecular forces become significant — the compressibility factor $Z = PV/nRT$ drifts above 1. At low temperature (or near a phase change), attractive intermolecular forces become comparable to KE, and $Z$ drops below 1. The flat region of a real-gas isotherm at low $T$ is condensation; an ideal gas would never show this.
Which gas sample has the most molecules from equal masses?
The gas with the smallest molar mass $M_r$. From $n = m/M_r$, equal mass means more moles for smaller $M_r$, and therefore more molecules ($N = n N_A$). For example, 1 g of H$_2$ contains roughly $6\times$ more molecules than 1 g of N$_2$. This is a recurring multiple-choice format in Paper 1.
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